Question

5 drops of 0.15 M Ki added to
40 drops of Na2S2O3
What is the final concentration of ki?​

Answer 1

`\boxed{\text{0.017 mol/L}}`

Step-by-step explanation:

Na₂S₂O₃ solution does not react with KI (it reacts with I₂), so it is simply diluting the KI, and we can use the dilution formula.

`c_(1)V_(1) = c_(2)V_(2)`

Data:

c₁ = 0.15 mol·L⁻¹; V₁ = 5 drops

V(Na₂S₂O₃) = 40 drops

Calculations:

(a) Calculate the total volume

V₂ = 5 + 40 = 45 drops

(b) Calculate the concentration

0.15 × 5 = c₂ × 45

0.75 = 45c₂

`c_(2) = (0.75 )/(45) = \boxed{\textbf{0.017 mol/L}}`